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2 years ago

Which of these is a chemical property of a substance

1 answer:

3 0

A chemical property is a characteristic of a substance that may be observed when it participates in a chemical reaction. Examples of chemical properties include flammability, toxicity, chemical stability, and heat of combustion

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A laboratory analysis of a sample finds it is composed of 38.8% carbon, 16.2% hydrogen, and 45.1% nitrogen. What is its empirica

Sladkaya [172]

Answer: The empirical formula for the given compound is $CH_5N$

Explanation : Given,

Percentage of C = 38.8 %

Percentage of H = 16.2 %

Percentage of N = 45.1 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 38.8 g

Mass of H = 16.2 g

Mass of N = 45.4 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{38.8g}{12g/mole}=3.23moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{16.2g}{1g/mole}=16.2moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{45.4g}{14g/mole}=3.24moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.23 moles.

For Carbon = $\frac{3.23}{3.23}=1$

For Hydrogen = $\frac{16.2}{3.23}=5.01\approx 5$

For Oxygen = $\frac{3.24}{3.23}=1.00\approx 1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 1 : 5 : 1

Hence, the empirical formula for the given compound is $C_1H_5N_1=CH_5N$

3 0

3 years ago

An element has five isotopes. Calculate the atomic mass of this element using the information below. Show all your work. Using t

Katena32 [7]

Answer: Sol:-

Data provided in the question is :-

Atomic mass of isotope -1 = 64 amu

Atomic mass of isotope -2 = 66 amu

Atomic mass of isotope -3 = 67 amu

Atomic mass of isotope -4 = 68 amu

Atomic mass of isotope - 5 = 70 amu

Percentage abundace of isotope - 1 = 48.89 %

Percentage abundance of isotope -2 = 27.81 %

Percentage abundance of isotope - 3 = 4.11%

Percentage abundance of isotope-4 = 18.57%

Percentage abundance of isotope - 5 = 0.62 %

Formula used :-

Average atomic mass of an element =[ {(atomic mass of isotope-1 * percentage abundance of isotope-1) + ( atomic mass of isotope-2 * percentage abundance of isotope -2) + ( atomic mass of isotope -3 * percantege abundance of isotope-3 ) + ( atomic mass of isotope-4 * percentage abundance of isotope-4) + (atomic mass of isotope-5 * percentage abundance of isotope-5)} / 100]

Calculation :-

Put all the value in the formula :-

Average atomic mass of an element = [{(64 * 48.89) + (66 * 27.81) + (67 * 4.11) + (68 * 18.57) + (70 * 0.62)} / 100] amu

= [{(3128.96) + (1835.46) +(257.37) + (1262.76) + (43.4)} / 100] amu

= {(6528.04) / 100} amu

= 65.2804 amu

Average atomic mass of an element is = 65.2804 amu

Then this mass is approximatly equal to atomic mass of zinc so this element would be zinc

atomic mass of zinc = 65.38 \approx 65.2804 amu

5 0

2 years ago

Balance the following redox reaction in acidic solution. Zn(s)+MnO−4(aq)→ Zn+2(aq)+Mn+2(aq)

devlian [24]

Answer:

2MnO₄⁻ + 5Zn + 16H⁺ → 2Mn²⁺ + 8H₂O + 5Zn²⁺

Explanation:

To balance a redox reaction in an acidic medium, we simply follow some rules:

Split the reaction into an oxidation and reduction half.
By inspecting, balance the half equations with respect to the charges and atoms.
In acidic medium, one atom of H₂O is used to balance up each oxygen atom and one H⁺ balances up each hydrogen atom on the deficient side of the equation.
Use electrons to balance the charges. Add the appropriate numbers of electrons the side with more charge and obtain a uniform charge on both sides.
Multiply both equations with appropriate factors to balance the electrons in the two half equations.
Add up the balanced half equations and cancel out any specie that occur on both sides.
Check to see if the charge and atoms are balanced.

Solution

Zn + MnO₄⁻ → Zn²⁺ + Mn²⁺

The half equations:

Zn → Zn²⁺ Oxidation half

MnO₄⁻ → Mn²⁺ Reduction half

Balancing of atoms(in acidic medium)

Zn → Zn²⁺

MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Balancing of charge

Zn → Zn²⁺ + 2e⁻

MnO₄⁻ + 8H⁺ + 5e⁻→ Mn²⁺ + 4H₂O

Balancing of electrons

Multiply the oxidation half by 5 and reduction half by 2:

5Zn → 5Zn²⁺ + 10e⁻

2MnO₄⁻ + 16H⁺ + 10e⁻→ 2Mn²⁺ + 8H₂O

Adding up the two equations gives:

5Zn + 2MnO₄⁻ + 16H⁺ + 10e⁻ → 5Zn²⁺ + 10e⁻ + 2Mn²⁺ + 8H₂O

The net equation gives:

5Zn + 2MnO₄⁻ + 16H⁺ → 5Zn²⁺ + 2Mn²⁺ + 8H₂O

8 0

3 years ago

A solution of potassium hydroxide reacts completely with a solution of nitric acid. What solid mixture does it create?

KIM [24]

Answer:

$\boxed{\text{KNO}_{3}}$

Explanation:

The reaction is

KOH(aq) + HNO₃(aq) ⟶ KNO₃(aq) + H₂O(ℓ)

If you evaporate the water, the solid substance is the compound, potassium nitrate.

$\boxed{\textbf{KNO}_{3}}$

KNO₃(aq) ⟶ KNO₃(s)

4 0

2 years ago

Iodine-131 is a radioactive isotope with a half-life of 8 days. how many grams of a 64 g sample of iodine-131 will remain at the

icang [17]

At the end of 24 days, there will only be 16 g.

7 0

2 years ago